One half of a kilogram of liquid water at 273 K (0 °C) is placed outside on a da
ID: 1612053 • Letter: O
Question
One half of a kilogram of liquid water at
273 K (0 °C)
is placed outside on a day when the temperature is
262 K (11 °C).
Assume that the heat is lost from the water only by means of radiation and that the emissivity of the radiating surface is 0.60. Consider two cases: when the surface area of the water is
0.032 m2
(as it might be in a cup) and
1.4 m2
(as it could be if the water were spilled out to form a thin sheet). (The latent heat of fusion of water is 33.5 104 J/kg.)
A)For the case of the water with a surface area of
0.032 m2,
determine the time it takes the water to freeze into ice at
0 °C in seconds
B)For the case of the water with a surface area of
1.4 m2,
determine the time it takes the water to freeze into ice at
0 °C in seconds
Explanation / Answer
A) The rate of heat lost, dQ/dt = sigma*e*A*(T1^4 - T2^4)
= 5.67*10^-8*0.6*0.032*(273^4 - 262^4)
= 0.9172 J/s or W
Heat must be removed o freeze the Ice, dQ = m*Lf
= 0.5*3.33*10^5
= 1.665*10^5 J
time taken, dt = dQ/(dQ/dt)
= 1.665*10^5/0.9172
= 1.815*10^5 s
B)
The rate of heat lost, dQ/dt = sigma*e*A*(T1^4 - T2^4)
= 5.67*10^-8*0.6*1.4*(273^4 - 262^4)
= 40.13 J/s or W
time taken, dt = dQ/(dQ/dt)
= 1.665*10^5/40.13
= 4.15*10^3 s
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